Dear users,
I have a matrix b as
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 1 1
[3,] 0 1 0 1
[4,] 0 1 1 2
[5,] 1 0 0 1
[6,] 1 0 1 2
[7,] 1 1 0 2
[8,] 1 1 1 3
with last column the rowSums(b)
I want for each value of last column of b separately (b:0,1,2,3, to sort
the above table by reverse lexicographical order.
For the above table the result should be
1 0 0 0
2 1 0 0
3 0 1 0
4 0 0 1
5 1 1 0
6 1 0 1
7 0 1 1
8 1 1 1
Could anyone help me with this?
Thanks alot
Evgenia
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Hi, see the code below. Next time, please provide self-contained code to
generate the matrix (read the posting guide) so that we can just copy-paste
it into R to get the matrix.
x=rep(c(0,1),4)
y=rep(c(0,0,1,1),2)
z=rep(c(0,1),each=4)
m=cbind(z,y,x)
w=rowSums(m)
m=cbind(m,w)
m
m[order(w,x,y,z),]
HTH,
Daniel
-------------------------
cuncta stricte discussurus
-------------------------
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On
Behalf Of Evgenia
Sent: Friday, February 19, 2010 12:10 PM
To: r-help at r-project.org
Subject: [R] inverse lexicographical ordering
Dear users,
I have a matrix b as
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 1 1
[3,] 0 1 0 1
[4,] 0 1 1 2
[5,] 1 0 0 1
[6,] 1 0 1 2
[7,] 1 1 0 2
[8,] 1 1 1 3
with last column the rowSums(b)
I want for each value of last column of b separately (b:0,1,2,3, to sort
the above table by reverse lexicographical order.
For the above table the result should be
1 0 0 0
2 1 0 0
3 0 1 0
4 0 0 1
5 1 1 0
6 1 0 1
7 0 1 1
8 1 1 1
Could anyone help me with this?
Thanks alot
Evgenia
--
View this message in context:
http://n4.nabble.com/inverse-lexicographical-ordering-tp1561930p1561930.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Thanks for your help.
Finally I found the solution to this problem for any matrix without having
to know its number of columns
I write it down in case someone else needed it.
#Creation of b matrix
ncol<-3
b <- matrix(0, 2^ncol, ncol)
for (i in 1:ncol)
b[, ncol+1-i] <- rep(rep(c(0,1),c(2^(i-1),2^(i-1))),2^(ncol-i))
b<-cbind(b,rowSums(b))
mat <- b[ ,ncol(b):1 ] ( reverse columns making last first, etc)
b[do.call("order", split(mat, col(mat))),]
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